Question

if x is equals to 6 minus 4 root 2 then find root x + 1 by root x ​

Answer 1

Answer:

$\sqrt{x} + \frac{1}{ \sqrt{x} } = 3 + \frac{1}{ \sqrt{2} } - \sqrt{2}$

Step-by-step explanation:

$x = 6 - 4 \sqrt{2} \\ x = ( \sqrt{2} )^{2} - (2 \times \sqrt{2} \times 2) + (2)^{2} \\ x = (2 - \sqrt{2} )^{2} \\ \sqrt{x} = 2 - \sqrt{2}$

hence,

$\frac{1}{ \sqrt{x} } = \frac{1}{2 - \sqrt{2} } \\ \frac{1}{ \sqrt{x} } = \frac{(2 + \sqrt{2} )}{(2 - \sqrt{2} ) \times (2 + \sqrt{2} )} \\ \frac{1}{ \sqrt{x} } = \frac{(2 + \sqrt{2} )}{2} \\ \frac{1}{ \sqrt{x} } = 1 + \frac{1}{ \sqrt{2} }$

Adding both we get

$\sqrt{x} + \frac{1}{ \sqrt{x} } = 3 + \frac{1}{ \sqrt{2} } - \sqrt{2}$

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